Question: $y^2-x^2y+3x^3=4$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2x-2y}{27x^2}$ (Choice B) B $\dfrac{2xy-9x^2}{2y-x^2}$ (Choice C) C $\dfrac{-27x^2}{2y-2x}$ (Choice D) D $\dfrac{2y+x^2}{2xy-9x^2}$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $y^2-x^2y+3x^3=4$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} y^2-x^2y+3x^3&=4 \\\\ \dfrac{d}{dx}(y^2-x^2y+3x^3)&=\dfrac{d}{dx}(4) \\\\ \dfrac{d}{dx}(y^2)-\dfrac{d}{dx}(x^2y)+\dfrac{d}{dx}(3x^3)&=0 \\\\ 2y\cdot\dfrac{dy}{dx}-\Bigl(2x\cdot y+x^2\cdot\dfrac{dy}{dx}\Bigr)+9x^2&=0 \\\\ 2y\cdot\dfrac{dy}{dx}-2xy-x^2\cdot\dfrac{dy}{dx}+9x^2&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 2y\cdot\dfrac{dy}{dx}-2xy-x^2\cdot\dfrac{dy}{dx}+9x^2&=0 \\\\ \dfrac{dy}{dx}(2y-x^2)&=2xy-9x^2 \\\\ \dfrac{dy}{dx}&=\dfrac{2xy-9x^2}{2y-x^2} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{2xy-9x^2}{2y-x^2}$.